Question

in an r-l-c series circuit, r=10 ; l=1h and c= 1f. it is connected to 230 v a.c. source of variable frequency. when the frequency is set to zero, circuit current will be

Answer 1

Answer:

(a)

The resonance frequency is given by

ω=

LC

1

=

5×80×10

−6

1

=50rad/s

The resonant frquency is 50 rad/s.

(b)

At resonance, ωL=1/ωC

⇒Z=R=40Ω

The peak voltage, V

o

=

2

V

I

o

=

2

V/Z

I

o

=8.13A

(c)

Potential drop across inductor,

V

L(rms)

=I×ω

R

L

I=I

0

/

2

=

2

V/

2

Z=23/40 A

⇒V

rms

=1437.5V

Potential drop across capacitor,

(V

c

)

rms

=I×

w

r

C

1

=1437.5V

Potential drop across resistor,

(V

R

)

rms

=I×R=230V

Potential drop across the LC combination

V

LC

= I(X

L

-X

C

)

at resonance X

L

=X

C

⇒V

LC

=0 V